Transistor Circuits I

Common-Base, DC operation

The humble transistor

Q1

Emitter (E)

Collector (C)

Base (B)

Transistor basics

• Emitter to base junction is forward biased (normally)

• Collector to base junction is reverse biased (normally)

• Transistors are current operated devices, so KCL should be applied first:

–IE = IC + IB

Basics continued

• Leakage current: ICBO (Emitter open)

– Usually is considered negligible, but can affect things when IC is small

Basics continued

• hFB = α = 𝐼𝐶

𝐼𝐸 (α ≤ 1)

• hFE = β = 𝐼𝐶

𝐼𝐵 (β has no definitive limit)

• The symbols hFB and hFE are BJT parameters; the h means Hybrid parameter. The subscript FB means Forward current transfer ratio of the common, or grounded, Base circuit, whereas FE means Forward current transfer ratio of the common, or grounded, Emitter circuit.

Basic configuration of Common-Base

First circuit

• If VEE = 20V and VEB is negligible, find IE when RE equals (a) 80kΩ, (b) 40kΩ, (c) 20kΩ, (d) 10kΩ, (e) 5kΩ, and (f) 1kΩ.

Work for first circuit

Formula is 𝐼𝐸 =𝑉𝐸𝐸

𝑅𝐸=

20V

𝑅𝐸

• (a) 𝐼𝐸 =20V

80kΩ= 0.25mA = 250µA;

• (b) 𝐼𝐸 =20V

40kΩ= 0.5mA = 500µA;

• (c) 𝐼𝐸 =20V

20kΩ= 1mA;

• (d) 𝐼𝐸 =20V

10kΩ= 2mA;

• (e) 𝐼𝐸 =20V

5kΩ= 4mA;

• (f) 𝐼𝐸 =20V

1kΩ= 20mA

Use approximation for VEB

If VEB = 0.7V, rework previous values to determine IE in each instance.

• Formula is now 𝐼𝐸 =𝑉𝐸𝐸−𝑉𝐸𝐵

𝑅𝐸=

20V−0.7V

𝑅𝐸=

19.3V

𝑅𝐸

• (a) 𝐼𝐸 =19.3V

80kΩ= 241.25µA;

• (b) 𝐼𝐸 =19.3V

40kΩ= 482.5µA;

• (c) 𝐼𝐸 =19.3V

20kΩ= 965µA;

• (d) 𝐼𝐸 =19.3V

10kΩ= 1.93mA;

• (e) 𝐼𝐸 =19.3V

5kΩ= 3.86mA;

• (f) 𝐼𝐸 =19.3V

1kΩ= 19.3mA

Proving the point (Measuring (a))

Measuring (b)

Measuring (c)

Measuring (d)

Measuring (e)

Measuring (f)

PNP versions

Second circuit

• If RE = 10kΩ and VEB = 0V, what are the values of IE, IC and VCB when RC equals (a) 5kΩ, (b) 8kΩ, (c) 10kΩ, (d) 12kΩ. Assume hFB = α = 1.

10kΩ

18V -22V

Points to ponder

• Since IE will remain constant, use the value found from calculating through the derivation

of Ohm’s Law: 𝐼𝐸 =𝑉𝐸𝐸

𝑅𝐸=

18V

10kΩ= 1.8mA.

Additionally, we are treating α as 1, which

means using the formula 𝐼𝐶

𝐼𝐸= 1 translates into

IE and IC being equal. Therefore, the value of IC for all changes of RC remains a constant 1.8mA (so long as saturation is avoided…)

Additional points

• Following KVL, VCC = VRC + VCB. Since VRC = ICRC (Ohm’s Law), substituting this into the original formula gives us VCC = ICRC + VCB. Reworking the equation to solve for VCB yields VCB = VCC – ICRC (You can use the absolute value for VCC when calculating, but remember to include polarity when writing the final answer).

Work through for circuit

• (a) 𝑉𝐶𝐵 = 𝑉𝐶𝐶 − 𝐼𝐶𝑅𝐶 = 22 − 1.8mA 5kΩ =22 − 9 = −13V (Remember the polarity of VCC is negative…)

• (b) 𝑉𝐶𝐵 = 𝑉𝐶𝐶 − 𝐼𝐶𝑅𝐶 = 22 − 1.8mA 8kΩ =22 − 14.4 = −7.6V

• (c) 𝑉𝐶𝐵 = 𝑉𝐶𝐶 − 𝐼𝐶𝑅𝐶 = 22 − 1.8mA 10kΩ =22 − 18 = −4V

• (d) 𝑉𝐶𝐵 = 𝑉𝐶𝐶 − 𝐼𝐶𝑅𝐶 = 22 − 1.8mA 12kΩ =22 − 21.6 = −0.4V

Third circuit

• If VCC = 24V, find (a) the saturation current, IC(sat). If VEB = 0.7V, what are the values of IE, IC, and VCB when RE equals (b) 20kΩ, (c) 10kΩ, and (d) 5kΩ. Assume αDC = 1.

IC(sat) and some constants in our calculations

• (a) 𝐼𝐶(𝑠𝑎𝑡) =𝑉𝐶𝐶

𝑅𝐶=

24V

5kΩ= 3mA

• VRE = VEE – VEB = 20 – 0.7 = 19.3V (technically this should be written as -19.3V since VEE is negative in polarity)

Calculating the rest

• (b) 𝐼𝐸 =𝑉𝑅𝐸

𝑅𝐸=

19.3V

20kΩ= 965µA ∴ 𝐼𝐶 = 965µA;

VCB = VCC – ICRC = 24 – (965µA)(8kΩ) = 24 – 7.72 = 16.28V

And the beat goes on…

• (c) 𝐼𝐸 =𝑉𝑅𝐸

𝑅𝐸=

19.3V

10kΩ= 1.93mA ∴ 𝐼𝐶 =

1.93mA; VCB = VCC – ICRC = 24 – (1.93mA)(8kΩ) = 24 – 15.44 = 8.56V

And on…

• (d) 𝐼𝐸 =𝑉𝑅𝐸

𝑅𝐸=

19.3V

5kΩ= 3.86mA ∴ 𝐼𝐶 =

3mA (this cannot be exceeded); VCB = VCC – ICRC = 24 – (3mA)(8kΩ) = 24 – 24 = 0V

Measure to make sure (a)

(b)

(c)

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